Meromorphic continuation of L-functions I am following these notes and on page 2 the claim is that if we have an $L$-function $$L(s) = \sum_{n=1}^{\infty}\frac{a_n}{n^s}$$  with $a_n=O(n^r)$ and if $L$ has a meromorphic continuation and satisfies a functional equation (both written explicitly on page 2 of the notes linked above) then we have the following integral expression: $$\hat{L}(s) = \int_1^{\infty} (x^s+wx^{(k-s)/2})\Theta(\sqrt{N}x) \frac{dx}{x}$$
where $\Theta(x)=\sum_{n=1}^{\infty} a_n \theta_{n, \gamma}(x)$ and the $\theta$ functions depend only on $\gamma$ (again defined on page 2 of the notes linked above).
Now I have a couple of questions about this. Is there a reference that has a complete proof of this claim? I couldn't find one but what I do have is a set of notes(see pages 18-21) which have a complete proof of the meromorphic continuation of the classical $\zeta$-function. I tried to follow the same argument as the one on page 18 to get the above integral expression but failed. More specifically, if I start the argument as on page 18 then I get the following expression: $$\int_0^{\infty} \left[\sum_{n=1}^\infty a_nf(nt) \right] t^{s-1}dt.$$ In the case of the $\zeta$-function, we have $a_n=1$ and we can apply the Poisson Summation Formula for the function $h(t)=f(nt)$ but here we have arbitrary $a_n$ and I don't know how to proceed.
Thank you for taking the time to read this and let me know if my question is not clear enough in any way!
 A: I believe there is some small error in the notes. I'll try to derive what the notes is getting at.
On your attempt
We indeed can't apply Poisson summation formula here due to the $a_n$, but it's also not necessary. Poisson summation is used to show symmetry of classical theta series $\sum_n e^{-\pi n^2x}$, whose role should be played by (*) in page 2 of your notes. (i.e. it comes from the functional equation of the $L$-function at play)

On deriving the symmetry of theta function
We first try to derive symmetry of "theta function" from the functional equation, which should be a consequence of inverse Mellin transforming the functional equation.
Details:
Let
$$\gamma(s) = \int_0^{\infty} \theta(t) t^s \frac{dt}{t} = n^s \int_0^{\infty} \theta(nt) t^s \frac{dt}{t}$$
Hence
$$n^{-s} \gamma(s) = \int_0^{\infty} \theta(nt) t^s \frac{dt}{t}$$
Weight each term by $a_n$, and sum over them - note that summation and integral can be interchanged due to fast exponential decay of $\theta(nt)$ -
$$L(s) \gamma(s) = \int_0^{\infty} \left(\sum_n a_n \theta(nt) \right) t^s \frac{dt}{t}$$
Hence
$$\hat{L}(s) = \int_0^{\infty} \left(\sum_n a_n \theta(nt) \right) \left(\frac{N^{1/2}}{\pi^{d/2}}t\right)^s \frac{dt}{t} = \int_0^{\infty} \left(\sum_n a_n \theta(\frac{nt \pi^{d/2}}{N^{1/2}})\right) t^s \frac{dt}{t}$$
Let's call the inner term $\Theta(t) = \sum_n a_n \theta(\frac{nt \pi^{d/2}}{N^{1/2}})$ - not exactly the $\Theta(t)$ in the notes, but close enough, then
$$\hat{L}(s) = \int_0^{\infty} \Theta(t) t^s \frac{dt}{t}$$
Similarly for the dual $L$-function $\bar{L}(s) = \sum \bar{a_n} n^{-s}$, we can form $\bar{\Theta}(t) = \sum_n \bar{a_n} \theta(\frac{nt \pi^{d/2}}{N^{1/2}})$, and
$$w \hat{\bar{L}}(k-s) = w \left(\frac{N}{\pi^d}\right)^{(k-s)/2} \gamma(k-s) \bar{L} (k-s) = w \int_0^{\infty} \bar{\Theta}(t) t^{k-s} \frac{dt}{t} = \int_0^{\infty} w \cdot t^{-k} \cdot \bar{\Theta} \left(\frac{1}{t}\right) t^s \frac{dt}{t}$$
The functional equation says
$$\hat{L} (s) = w \hat{\bar{L}}(k-s)$$
so
$$\int_0^{\infty} \Theta(t) t^s \frac{dt}{t} = \int_0^{\infty} w \cdot t^{-k} \cdot \bar{\Theta} \left(\frac{1}{t}\right) t^s \frac{dt}{t}$$
The analytic conditions (a) of $L$ series should allow inverse Mellin transform to apply, so that you can write
$$\Theta(t) = w \cdot t^{-k} \cdot \bar{\Theta} \left(\frac{1}{t}\right)$$

On deriving the integral formula
From here, you can follow the recipe of Riemann zeta, it would look something like this.
$$\hat{L}(s) = \int_0^{\infty} \Theta(t) t^s \frac{dt}{t}$$
Split the integral to $[0,1]$ and $[1, \infty)$, and do a coordinate change $t \to \frac{1}{t}$ for $[0,1]$, then
$$\hat{L}(s) = \int_1^{\infty} \Theta\left(\frac{1}{t}\right) t^{-s} \frac{dt}{t} + \int_1^{\infty} \Theta(t) t^{s} \frac{dt}{t}$$
Applying the functional equation to the first integral,
$$\hat{L}(s) = \int_1^{\infty} \left(w \bar{\Theta}(t) t^{k-s} + \Theta(t) t^s\right) \frac{dt}{t}$$
This formula is likely what you are looking for.
There are a few differences with the one in the notes, but I think this one is right:

*

*if the $L$-function is not self dual, i.e. $a_n \not\in \mathbb{R}$ or i.e. $L(s) \neq \bar{L}(s)$, it's not clear to me how you can merge the $\bar{\Theta}$ and $\Theta$ term.

*If you'd like, you can probably do one more substitution $t \to t^{1/2}$ to match the $s/2$ and $(k-s)/2$ exponents, but that's probably unnecessary.

*You can also do another $t \to N^{1/2}t$ substitution to remove dependency of $\Theta$ on $N$, and that would give you the extra $\sqrt{N}$ in the final integral.

